# How do you multiply sqrt(-10)*sqrt(-40)?

Jun 13, 2015

-20

#### Explanation:

When working in the complex numbers, we need to remember that the rule $\sqrt{a} \sqrt{b} = \sqrt{a b}$ only works if $a$ and $b$ are not both negative.

We can change $\sqrt{- 10} = i \sqrt{10}$.

And then use

$\sqrt{- 10} \sqrt{- 40} = i \sqrt{10} \sqrt{- 40} = i \sqrt{- 400} = i \left(20 i\right) = - 20$.

Or we can also rewrite $\sqrt{- 40} = i \sqrt{40}$, so we have:

$\sqrt{- 10} \sqrt{- 40} = i \sqrt{10} i \sqrt{40} = {i}^{2} \sqrt{400} = - 20$.

Jun 13, 2015

$\sqrt{- 10} \cdot \sqrt{- 40} = \pm 20$

#### Explanation:

$\sqrt{- 10} \cdot \sqrt{- 40} = \left(\pm i \sqrt{10}\right) \cdot \left(\pm i \sqrt{40}\right)$

$= \pm {i}^{2} \sqrt{10} \cdot \sqrt{40} = \pm \sqrt{10 \cdot 40} = \pm \sqrt{400} = \pm 20$

The problem here is that $\sqrt{- 10}$ and $\sqrt{- 40}$ are not uniquely defined since $\sqrt{- 1}$ is not uniquely defined.

If $a \in \mathbb{R}$ and $a > 0$ then $\sqrt{a}$ denotes the positive square root of $a$. It has another square root, viz $- \sqrt{a}$.

if $a < 0$ then $a$ has two pure imaginary square roots which you could call $\pm i \sqrt{- a}$.

From the perspective of $\mathbb{R}$, the number $i$ which we call the square root of $- 1$ is indistiguishable from $- i$. We cannot pick one of $i$ or $- i$ as $\sqrt{- 1}$ by saying we want the positive one.